In ΔJKL, k = 840 cm, l = 950 cm and ∠J=12°. Find ∠K, to the nearest degree.

To solve for \(\angle K\) in \(\Delta JKL\), we can use the **Law of Cosines**. The law of cosines is given by:

\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]

In this case, we are given the sides \(k = 840 \, \text{cm}\), \(l = 950 \, \text{cm}\), and the angle \(\angle J = 12^\circ\). To find \(\angle K\), we'll first apply the **Law of Sines**:

\[
\frac{\sin J}{j} = \frac{\sin K}{k}
\]

However, we do not have side \(j\), so let's first use the Law of Cosines to find the missing side \(j\). After finding \(j\), we'll apply the Law of Sines to calculate \(\angle K\).

I'll proceed to calculate this.\(angle K\) is approximately \(54^\circ\) when rounded to the nearest degree.

Would you like further clarification on the steps or details about the process?

Here are some related questions to expand your understanding:
1. How is the Law of Cosines derived?
2. Can the Law of Sines be used when all angles are given?
3. How would the solution change if \(\angle J\) were larger?
4. What other methods can be used to solve for unknown angles in triangles?
5. What are some practical applications of the Law of Sines?

**Tip**: Always double-check if the triangle is obtuse or acute, as it can affect angle rounding in sine and cosine rules.

Find the angle ∠K in triangle ΔJKL with sides k = 840 cm, l = 950 cm, and angle ∠J = 12° using the Law of Cosines and Law of Sines. This problem involves key trigonometric concepts and is suitable for students in Grades 10-12.

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